如何证明？

王守恩

       \(\sin\left(\arccos(\frac{\sin\theta}{2})\right)=\frac{\sqrt{3(\sin\theta)^2+4(\cos\theta)^2}}{2}\)

       \(\sin\left(\arccos(\frac{\cos\theta}{2})\right)=\frac{\sqrt{4(\sin\theta)^2+3(\cos\theta)^2}}{2}\)

       \(\cos\left(\arcsin(\frac{\sin\theta}{2})\right)=\frac{\sqrt{3(\sin\theta)^2+4(\cos\theta)^2}}{2}\)

       \(\cos\left(\arcsin(\frac{\cos\theta}{2})\right)=\frac{\sqrt{4(\sin\theta)^2+3(\cos\theta)^2}}{2}\)

王守恩

求证：
\(\D\left(\sin(\arccos(\frac{\sin\theta}{3}))-\frac{\cos\theta}{3}\right)\left(\sin(\arccos(\frac{\sin\theta}{3}))+\frac{\cos\theta}{3}\right)=\frac{8}{9}\)

葡萄糖

\begin{align*}
\sin\left(\arcsin\left(t\right)\right)&=\qquad\!\!t&
\cos\left(\arcsin\left(t\right)\right)&=\sqrt{1-t^2}&
\tan\left(\arcsin\left(t\right)\right)&=\dfrac{t}{\sqrt{1-t^2}}
\\\\
\sin\left(\arccos\left(t\right)\right)&=\sqrt{1-t^2}&
\cos\left(\arccos\left(t\right)\right)&=\qquad\!\!t&
\tan\left(\arccos\left(t\right)\right)&=\dfrac{\sqrt{1-t^2}}{t}
\\\\
\sin\left(\arctan\left(t\right)\right)&=\dfrac{t}{\sqrt{1+t^2}}&
\cos\left(\arctan\left(t\right)\right)&=\dfrac{1}{\sqrt{1+t^2}}&
\tan\left(\arctan\left(t\right)\right)&=\qquad\!\!t
\\\\
\sin\left(\operatorname{arccot}\left(t\right)\right)&=\dfrac{1}{\sqrt{1+t^2}}&
\cos\left(\operatorname{arccot}\left(t\right)\right)&=\dfrac{t}{\sqrt{1+t^2}}&
\tan\left(\operatorname{arccot}\left(t\right)\right)&=\qquad\!\!\dfrac{1}{t}
\\\\
\sin\left(\operatorname{arcsec}\left(t\right)\right)&=\dfrac{\sqrt{t^2-1}}{t}&
\cos\left(\operatorname{arcsec}\left(t\right)\right)&=\qquad\!\!\dfrac{1}{t}&
\tan\left(\operatorname{arcsec}\left(t\right)\right)&=\sqrt{t^2-1}&
\\\\
\sin\left(\operatorname{arccsc}\left(t\right)\right)&=\qquad\!\!\dfrac{1}{t}
&\cos\left(\operatorname{arccsc}\left(t\right)\right)&=\dfrac{\sqrt{t^2-1}}{t}
&\tan\left(\operatorname{arccsc}\left(t\right)\right)&=\dfrac{1}{\sqrt{t^2-1}}\\
\end{align*}

王守恩

葡萄糖 发表于 2019-8-8 12:38
\begin{align*}
\sin\left(\arcsin\left(t\right)\right)&=\qquad\!\!t&
\cos\left(\arcsin\left(t\right ...

       设   \(\arcsin(\frac{\cos\theta}{3})=\psi，-\frac{\pi}{2}<\psi<\frac{\pi}{2}\)

       这时必有   \(\sin\psi=\sin(\arcsin(\frac{\cos\theta}{3}))=\frac{\cos\theta}{3}\)

       \(\left(\cos(\arcsin(\frac{\cos\theta}{3}))-\frac{\sin\theta}{3}\right)\left(\cos(\arcsin(\frac{\cos\theta}{3}))+\frac{\sin\theta}{3}\right)\)

       \(=(\cos\psi-\frac{\sin\theta}{3})(\cos\psi+\frac{\sin\theta}{3})=(\cos\psi)^2-(\frac{\sin\theta}{3})^2\)

       \(=1-(\sin\psi)^2-(\frac{\sin\theta}{3})^2=1-(\frac{\cos\theta}{3})^2-(\frac{\sin\theta}{3})^2\)

       \(=1-\frac{(\cos\theta)^2}{9}-\frac{(\sin\theta)^2}{9}=1-\frac{(\cos\theta)^2+(\sin\theta)^2}{9}\)

       \(=1-\frac{1}{9}=\frac{8}{9}\)

王守恩

葡萄糖 发表于 2019-8-8 12:38
\begin{align*}
\sin\left(\arcsin\left(t\right)\right)&=\qquad\!\!t&
\cos\left(\arcsin\left(t\right ...

       \(\frac{\cos\theta}{3}=S(1)\)

       \(\arcsin(\frac{\cos\theta}{3})=S(2)\)

       \(\cos(\arcsin(\frac{\cos\theta}{3}))=S(3)\)

       \(\arcsin(\cos(\arcsin(\frac{\cos\theta}{3})))=S(4)\)

       \(\cos(\arcsin(\cos(\arcsin(\frac{\cos\theta}{3}))))=S(5)\)

       \(\cdots\cdots\)

       问：\(S(2019)=？\)

王守恩

王守恩 发表于 2019-8-8 17:22
\(\frac{\cos\theta}{3}=S(1)\)

       \(\arcsin(\frac{\cos\theta}{3})=S(2)\)

S(1)=S(5)=S(9)
S(2)=S(6)=S(10)
S(3)=S(7)=S(11)
S(4)=S(8)=S(12)

王守恩

王守恩 发表于 2019-8-8 17:22
\(\frac{\cos\theta}{3}=S(1)\)

       \(\arcsin(\frac{\cos\theta}{3})=S(2)\)

   \(n\) 是正整数
   
\(\D\left(\sin(\arccos(\frac{\sin\theta}{n}))\right)^2+\left(\frac{\sin\theta}{n}\right)^2=1\)  

\(\D\left(\cos(\arcsin(\frac{\cos\theta}{n}))\right)^2+\left(\frac{\cos\theta}{n}\right)^2=1\)